I have the following sequence I'm having trouble calculating the limit of. $$\lim_{n\to\infty}\frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times\cdots\times(2n)}$$ I got to the point where I found out this is a monotonic decreasing sequence in the range (0,1) I know it converges to 0, I'm just having trouble proving it. Any help would be great!!
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Does this answer your question? Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$ – Matcha Latte Jan 23 '21 at 21:01
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You can multiply numerator and denominator by $2\times 4\times 6\times\cdots\times (2n)$ to get $$\frac{1\times 2\times 3\times\cdots \times (2n-1)\times 2n}{(2\times 4\times 6\cdots \times 2n)^2}=\frac{(2n)!}{4^nn!n!}=\frac{1}{4^n}{2n\choose n}$$
Now you need to estimate ${2n\choose n}$. You can do this via Stirling's formula, or directly using bounds on the Central Binomial Coefficients. Your intuition is correct, the limit is zero.
vadim123
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this is the idea i have got at first sight.. not the stirlings formula and all but the idea of multiplying numerator and denominator by that number.. but then i had no idea what to do after that.. I would be thankful If you can let me know if there any other way to do from this other than this stirlings formula... – Nov 22 '13 at 15:24
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Hint:
Prove by induction that
$$x_n=\dfrac{1\cdot{3}\cdot\ldots\cdot{(2n-1)}}{2\cdot{4}\cdot\ldots\cdot{(2n)}}<\dfrac{1}{\sqrt{2n+1}}.$$
M. Strochyk
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If $$P=\frac{\prod_{k=1}^N(2k-1)}{\prod_{k=1}^N2k}$$ $$P=\frac{\Gamma(N+1/2)}{\sqrt\pi \Gamma(N+1)}$$ $$\lim_{N\to\infty}(P)=0$$
Riccardo.Alestra
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